Problem: Suppose we have a surface $S$ defined by the transformation $T$ for $0 < u < \pi$ and $0 < v < 2\pi$. $T(u, v) = (\cos(v), \cos(u) + v, \sin(v))$ What is the surface area of $S$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $6\pi$ (Choice B) B $2\pi$ (Choice C) C $\pi$ (Choice D) D $4\pi$
Explanation: Assume we have a surface $S$ parameterized by a transformation $T$. If we want to find the surface integral over $S$ of a function $f$, we can use the formula below to convert it into a familiar double integral. $ \iint_S f(T(u, v)) | T_u \times T_v | \, du \, dv$ Finding surface area using a surface integral means using $f(x, y, z) = 1$. In effect, we are saying that we only care about the scaling factor caused by the area element. Therefore: $A = \int_0^{2\pi} \int_0^{\pi} |T_u \times T_v| \, du \, dv$ Now we need to find the magnitude of the area element. $|T_u \times T_v| = |\sin(u)|$ [Calculation] The final step is to evaluate the double integral. We can substitute $\sin(u)$ for $|\sin(u)|$ because $\sin(u) > 0$ in the interval $0 < u < \pi$. $\begin{aligned} A &= \int_0^{2\pi} \int_0^{\pi} |T_u \times T_v| \, du \, dv \\ \\ &= \int_0^{2\pi} \int_0^{\pi} |\sin(u)| \, du \, dv \\ \\ &= \int_0^{2\pi} \int_0^{\pi} \sin(u) \, du \, dv \\ \\ &= \int_0^{2\pi} \left[ -\cos(u) \right]_0^{\pi} \, dv \\ \\ &= \int_0^{2\pi} 2 \, dv \\ \\ &= 4\pi \end{aligned}$ In conclusion, the surface area of $S$ is $4\pi$.